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Let $G$ is an abelian profinite group and $G=\varprojlim G_i$ (all $G_i$ are finite).

Then why $\hom(\varprojlim G_i,\mathbb{Q}/\mathbb{Z}) = \varinjlim\hom(G_i,\mathbb{Q}/\mathbb{Z})$ as topological groups.

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  • $\begingroup$ That's not true even as groups, you probably confused $\varinjlim$ and $\varprojlim$ $\endgroup$ – Max Jul 21 at 16:37
  • $\begingroup$ I may be wrong. Can you check profinite groups by Luis Ribes(Lemma 2.9.3)? $\endgroup$ – math Jul 21 at 19:31
  • $\begingroup$ check the answer(math.stackexchange.com/a/127291/326275). $\endgroup$ – math Jul 21 at 19:58
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    $\begingroup$ I assume that by $\operatorname{hom}$, you mean the group of continuous homomorphism, isn't it ? Then look for the compact subgroups of $\mathbb{Q/Z}$. Or directly, show that a continuous homomorphism $G\to\mathbb{Q/Z}$ has an open kernel. $\endgroup$ – Roland Jul 21 at 20:00
  • $\begingroup$ yes, a group of continuous homomorphism. $\endgroup$ – math Jul 21 at 20:07

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